If x,y,z∈ R+ and 16(16 x2 + y2 - 4 x y)=z(.16x+4y-z.), then

Question

If x,y,z∈ R+ and 16(16 x2 + y2 - 4 x y)=z(.16x+4y-z.), then (A) y,z,x are in A.P. (B) y,z,x are in G.P. (C) x,y,z are in A.P. (D) x,y,z are in G.P.

Tardigrade Admin · Accepted Answer

(16 x)2+(4 y)2+(z)2-(16 x)(4 y)-(4 y)(z-(z)(16 x)=0 ⇒ (1/2)(16 x-4 y)2+(4 y-z)2+(z-16 x)2 =0 ⇒ 16 x=4 y=z

Q. If $x, y, z \in R^{+}$ and $16 (16 x^{2} + y^{2} - 4 x y) = z (16 x + 4 y - z),$ then